I'm having trouble computing the integral $\int_{A}\log(\sin(x-y))dx dy$, where $A$ is the triangle defined by the 3 eqations: $x = \pi, y = x ,y = 0$

I tried the substition $u = x-y, v = x+y$ and got:

$$\frac{-1}{2}\int_{0}^{2\pi}\int_{0}^{v}\log(\sin u))du dv$$

I'm quite new at this, so there's a possibility I've made a silly mistake, or I'm just not getting something obvious about how to continue.

Any help would be appreciated!


The integral you are dealing with is: $$\int_0^{\pi} \int_0^x \ln(\sin(x-y))\,dy\,dx=\int_0^{\pi} \int_0^x \ln(\sin y)\,dy\,dx$$ Changing the order of integration: $$\int_0^{\pi} \int_y^{\pi} \ln(\sin y)\,dx\,dy=\int_0^{\pi} (\pi-y)\ln(\sin y)\,dy=I$$ $I$ is equivalent to: $$I=\int_0^{\pi}y\ln(\sin y)\,dy$$ Add the two expressions for $I$ and you get: $$I=\frac{\pi}{2}\int_0^{\pi} \ln(\sin y)\,dy=\boxed{-\dfrac{\pi^2}{2}\ln 2}$$

  • $\begingroup$ Thanks for the answer. How is the first equation justified? i.e., why does $sin(x-y)$ become $sin(y)$? Also, what allows you to change the order of integration? I thought that Fubini's theroem only allows this in the case of a rectangle domain. $\endgroup$ – Robert Jun 21 '14 at 10:49
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    $\begingroup$ @Robert: I used the property of definite integrals which states that: $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ About your second question, I am not sure what you ask. Do you mean changing the order of integration is invalid here? I am not sure about the correct terminology in double integrals. $\endgroup$ – Pranav Arora Jun 21 '14 at 10:58
  • $\begingroup$ Why is it possible to integrate by x and then by y instead of by y and then by x? I know that in that case of a rectangle domain (not a triangle like here) it is possible to switch the order of integration by Fubini's theorem. Why is it justified when the domain is a triangle? $\endgroup$ – Robert Jun 21 '14 at 11:04
  • $\begingroup$ @Robert: I didn't know if there was a theorem governing the change of order of integration in double integrals. I did it based on my intuition. I am sorry, I can't help with that. :( $\endgroup$ – Pranav Arora Jun 21 '14 at 11:07
  • $\begingroup$ You're probably correct anyway, thanks for the answer! $\endgroup$ – Robert Jun 21 '14 at 11:15

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